Shown in the figure is a shell made of a conductor. It has inner radius a and outer radius b and car

Q: Why is the charge on the inner surface not uniform?

The dipole creates a non-uniform electric field. Its positive end pulls electrons closer, while its negative end pushes them away. This creates a localized, non-uniform charge distribution on the inner surface to perfectly cancel the dipole's field inside the conductor.

Q: Does the dipole's field leak outside the shell?

Not at all! The induced charges on the inner surface perfectly cancel the dipole's electric field within the conducting material. This is called electrostatic shielding. The outside world only 'sees' the uniform charge on the outer surface.

Q: Why is the net charge on the inner surface exactly zero?

By Gauss's Law, the total enclosed charge within the conducting material must be zero. Since the dipole itself has a net charge of zero ($+q$ and $-q$), the inner surface must also have a net charge of zero to keep the total enclosed charge at zero.

Q: What if the shell was made of an insulating material instead of a conductor?

If it were an insulator, charges couldn't move freely to cancel the internal fields. The dipole's electric field would penetrate through the shell, and the field outside would be a complex superposition of the dipole's field and the shell's field.

Q: Why does the outer surface behave exactly like a point charge?

Because the outer surface has a perfectly uniform spherical charge distribution. According to Shell Theorem (which is derived from Gauss's Law), any spherically symmetric charge distribution creates an electric field outside that is mathematically identical to a point charge located at its center.

Question

Shown in the figure is a shell made of a conductor. It has inner radius $a$ and outer radius $b$ and carries charge $Q$. At its centre is a dipole $\mathbf{p}$ as shown.
In this case,

Options:

(A) surface charge density on the inner surface is uniform and equal to $\frac{Q/2}{4\pi a^2}$

(B) electric field outside the shell is the same as that of a point charge at the centre of the shell

(C) surface charge density on the outer surface depends on $|\mathbf{p}|$

(D) surface charge density on the inner surface of the shell is zero everywhere

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing inside a perfectly conducting spherical shell. This shell has an inner radius $a$ and an outer radius $b$.

The problem states that the shell itself carries a net charge $Q$.

Right at the very center of this hollow space, we place an electric dipole $\mathbf{p}$.

Our mission is to figure out how the charges distribute themselves on the inner and outer surfaces of the shell, and what the electric field looks like outside.

### The Magic of Gauss's Law in Conductors

To unlock this problem, we must rely on one of the most powerful principles in electrostatics: **the electric field inside the solid material of a conductor in electrostatic equilibrium is always exactly zero**.

If the field were not zero, the free electrons inside the conductor would move until they canceled it out.

Let's use this to our advantage. Imagine drawing a spherical Gaussian surface entirely within the solid conducting material, somewhere between radius $a$ and radius $b$.

According to Gauss's Law, the electric flux through this surface is proportional to the enclosed charge:

$$ \oint \mathbf{E} \cdot d\mathbf{A} = \frac{q_{	ext{enclosed}}}{\varepsilon_0} $$

Since $\mathbf{E} = 0$ everywhere on our Gaussian surface, the flux is zero. Therefore, the **net enclosed charge must be exactly zero**.

### The Inner Surface: A Tale of Induced Charges

Let's look closely at what is trapped inside our Gaussian surface. It encloses the central dipole and the inner surface of the conducting shell.

We know that an electric dipole consists of two equal and opposite charges ($+q$ and $-q$). Thus, the net charge of the dipole itself is zero.

For the total enclosed charge to be zero, the net induced charge on the inner surface of the shell must also be zero:

$$ q_{	ext{enclosed}} = q_{	ext{dipole}} + q_{	ext{inner\_surface}} = 0 $$

Since $q_{	ext{dipole}} = 0$, it immediately follows that $q_{	ext{inner\_surface}} = 0$.

However, this does not mean the inner surface is completely devoid of charge! The positive end of the dipole pulls electrons towards it, while the negative end pushes electrons away.

This creates a **non-uniform charge distribution** on the inner surface. The local charge density $\sigma_{	ext{inner}}$ varies from point to point, perfectly arranging itself to cancel the dipole's field inside the conductor.

### The Outer Surface: Perfect Symmetry

Now, let's shift our focus to the outer surface of the shell.

The problem tells us that the entire conducting shell carries a net charge $Q$. By the principle of conservation of charge, the sum of the charges on the inner and outer surfaces must equal this total charge:

$$ q_{	ext{total}} = q_{	ext{inner\_surface}} + q_{	ext{outer\_surface}} = Q $$

Since we just proved that the net charge on the inner surface is zero, the entire charge $Q$ must reside on the outer surface!

$$ q_{	ext{outer\_surface}} = Q $$

Because the electric field inside the conducting material is zero, the outer surface is completely shielded from the chaotic, non-uniform field of the dipole inside.

The outer surface only "knows" that it is a sphere and that it needs to distribute its charge $Q$ to minimize potential energy. Therefore, it spreads the charge perfectly uniformly.

The surface charge density on the outer surface is constant:

$$ \sigma_{	ext{outer}} = \frac{Q}{4\pi b^2} $$

### The Grand Conclusion

Finally, what does the electric field look like outside the shell?

Let's draw a massive Gaussian surface with a radius $r > b$ that completely engulfs the entire setup.

The total charge enclosed by this giant sphere is simply the net charge of the shell ($Q$) plus the net charge of the dipole ($0$). So, the total enclosed charge is just $Q$.

By Gauss's Law, a uniformly charged spherical surface behaves mathematically as if all its charge were concentrated at its center.

Therefore, the electric field outside the shell is identical to that of a single point charge $Q$ located at the origin:

$$ \mathbf{E} = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r^2} \hat{\mathbf{r}} $$

This beautiful phenomenon is known as **electrostatic shielding**. The outside world is completely oblivious to the complex dipole dynamics happening inside the cavity!

Animated Solution for Physics - Electrostatics: Shown in the figure is a shell made of a conductor. It has inner radius $a$ and outer radius $b$ and carries charge $Q$ . At its centre is a dipole $p$ as shown. In this case,

Select Answer:

Visualized Solution

Analyzing the Setup

Gauss's Law in a Conductor

Charge on the Inner Surface

Non-Uniform Inner Charge

Charge on the Outer Surface

Uniform Outer Charge Distribution

Electric Field Outside the Shell

The Way Forward

The Sigma Insight: Conductors

Analyzing the Setup

The Magic of Gauss's Law in Conductors

The Inner Surface

The Outer Surface

The Grand Conclusion

Similar Questions

A spherical metal shell $A$ of radius $R_{A}$ and a solid metal sphere $B$ of radius $R_{B} (< R_{A})$ are kept far apart and each is given charge $+ Q$ . Now they are connected by a thin metal wire. Then

An elliptical cavity is carved within a perfect conductor. A positive charge $q$ is placed at the centre of the cavity. The points $A$ and $B$ are on the cavity surface as shown in the figure. Then

A metallic shell has a point charge $q$ kept inside its cavity. Which one of the following diagrams correctly represents the electric lines of forces ?

Consider a neutral conducting sphere. A positive point charge is placed outside the sphere. The net charge on the sphere is then

Consider a neutral conducting sphere. A positive point charge is placed outside the sphere. The net charge on the sphere is then

A metallic solid sphere is placed in a uniform electric field. The lines of force follow the path(s) shown in figure as

Select Answer:

Visualized Solution

Analyzing the Setup

Gauss's Law in a Conductor

Charge on the Inner Surface

Non-Uniform Inner Charge

Charge on the Outer Surface

Uniform Outer Charge Distribution

Electric Field Outside the Shell

The Way Forward

The Sigma Insight: Conductors

Analyzing the Setup

The Magic of Gauss's Law in Conductors

The Inner Surface

The Outer Surface

The Grand Conclusion

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A metallic shell has a point charge q kept inside its cavity. Which one of the following diagrams correctly represents the electric lines of forces ?

Consider a neutral conducting sphere. A positive point charge is placed outside the sphere. The net charge on the sphere is then

Consider a neutral conducting sphere. A positive point charge is placed outside the sphere. The net charge on the sphere is then

A metallic solid sphere is placed in a uniform electric field. The lines of force follow the path(s) shown in figure as

A metallic shell has a point charge $q$ kept inside its cavity. Which one of the following diagrams correctly represents the electric lines of forces ?